擅长:python、mysql、java
<p>以下是一个不依赖外部包的解决方案:</p>
<pre><code>list = [1,1,1,1,1,1,2,3,4,4,5,1,2]
L = list + [999] # append a unique dummy element to properly handle -1 index
[l for i, l in enumerate(L) if l != L[i - 1]][:-1] # drop the dummy element
</code></pre>
<p>然后我注意到Ulf Aslak的类似解决方案更干净:)</p>